Calculate Ecell for the following electrochemical cell at 25 ºCPt (s) | H2 (g, 1.00 atm) | H+ (aq, 1.00 M) || Sn2+ (aq, 0.350 M) | Sn4+ (aq, 0.020 M) | Pt (s)The standard reduction potentials are as follows:Sn4+ (aq) + 2 e–à Sn2+ (aq) Eº = +0.13 V2 H+ (aq) + 2 e–à H2 (g) Eº = 0.00 V

The electrode potential of the cell is 0.093 V

Explanation:

The given chemical cell follows:

Oxidation half reaction:

Reduction half reaction:

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate the EMF of the cell, we use the Nernst equation, which is:

where,

= electrode potential of the cell = ? V

= standard electrode potential of the cell = +0.13 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

Hence, the electrode potential of the cell is 0.093 V

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8